<h2>Answer with explanation:-</h2>
Let
be the population mean .
By observing the given information, we have :-
![H_0:\mu=7450\\\\H_a:\mu\neq7450](https://tex.z-dn.net/?f=H_0%3A%5Cmu%3D7450%5C%5C%5C%5CH_a%3A%5Cmu%5Cneq7450)
Since the alternative hypotheses is two tailed so the test is a two-tailed test.
We assume that the life of a large shipment of CFLs is normally distributed.
(a) Given : Sample size : n=81 , since n>30 so we use z-test.
Sample mean : ![\overline{x}=7240](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D7240)
Standard deviation : ![\sigma=1350](https://tex.z-dn.net/?f=%5Csigma%3D1350)
Test statistic for population mean :-
![z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![\Rightarrow\ z=\dfrac{7240-7450}{\dfrac{1350}{\sqrt{81}}}\\\\\Rightarrow\ z=-1.4](https://tex.z-dn.net/?f=%5CRightarrow%5C%20z%3D%5Cdfrac%7B7240-7450%7D%7B%5Cdfrac%7B1350%7D%7B%5Csqrt%7B81%7D%7D%7D%5C%5C%5C%5C%5CRightarrow%5C%20z%3D-1.4)
The critical value (two-tailed) corresponds to the given significance level :-
![z_{\alpha/2}=z_{0.025}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3Dz_%7B0.025%7D%3D1.96)
Since the observed value of z (-1.4) is less than the critical value (1.96) , so we do not reject the null hypothesis.
Hence, we conclude that we have enough evidence to accept that the mean life is different from 7450 hours .
(b) The p-value :
, it means that the probability that the life of CFLs less than 7240 and greater than 7240 is 0.1615.
(c) The confidence interval for population mean is given by :-
![\overline{x} \pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=7240\pm(1.96)\dfrac{1350}{\sqrt{81}}\\\\=7240\pm294\\\\=(6946,7534)](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%5Cpm%5C%20z_%7B%5Calpha%2F2%7D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D7240%5Cpm%281.96%29%5Cdfrac%7B1350%7D%7B%5Csqrt%7B81%7D%7D%5C%5C%5C%5C%3D7240%5Cpm294%5C%5C%5C%5C%3D%286946%2C7534%29)
,